How Do You Calculate the Length of the Polar Curve \( r = 3 \sin \theta \)?

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SUMMARY

The discussion focuses on calculating the arc length of the polar curve defined by the equation \( r = 3 \sin \theta \) over the interval \( 0 \leq \theta \leq \frac{\pi}{3} \). The arc length formula used is \( \int \sqrt{r^{2} + \left(\frac{dr}{d\theta}\right)^{2}} d\theta \). The participants derive \( r^{2} = 9 \sin^{2}(\theta) \) and \( \frac{dr}{d\theta} = 3 \cos(\theta) \), leading to the integral \( \int \sqrt{9 \sin^{2}(\theta) + 9 \cos^{2}(\theta)} d\theta \). The integration process remains a challenge for the contributors, indicating a need for further clarification on integration techniques.

PREREQUISITES
  • Understanding of polar coordinates and polar curves
  • Familiarity with the arc length formula in calculus
  • Knowledge of differentiation and integration techniques
  • Basic trigonometric identities and their applications
NEXT STEPS
  • Review integration techniques for trigonometric functions
  • Study the application of the arc length formula in polar coordinates
  • Explore examples of polar curves and their properties
  • Learn about numerical integration methods for complex integrals
USEFUL FOR

Students studying calculus, particularly those focusing on polar coordinates and arc length calculations, as well as educators seeking to clarify integration techniques in this context.

Exeneva
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Homework Statement


r = 3 sin \vartheta

0 \leq \vartheta \leq \pi/3


Homework Equations



Arc Length: \int \sqrt{r^{2} + (dr/d\vartheta)^{2}}d\vartheta


The Attempt at a Solution


<br /> r^{2} = 9 (sin \vartheta)^{2} = 9 (1/2 - cos 2\vartheta/2)

r^{2} = 9/2 - 9/2 cos 2\vartheta

dr/d\vartheta = 3 cos \vartheta

\int (9/2 - 9/2 cos 2\vartheta + 3 cos \vartheta)^{1/2} d\vartheta
from 0 to \pi/3

Not sure how to integrate this.
 
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Exeneva said:

Homework Statement


r = 3 sin \vartheta

0 \leq \vartheta \leq \pi/3


Homework Equations



Arc Length: \int \sqrt{r^{2} + (dr/d\vartheta)^{2}}d\vartheta


The Attempt at a Solution


<br /> r^{2} = 9 (sin \vartheta)^{2} = 9 (1/2 - cos 2\vartheta/2)

r^{2} = 9/2 - 9/2 cos 2\vartheta

dr/d\vartheta = 3 cos \vartheta

\int (9/2 - 9/2 cos 2\vartheta + 3 cos \vartheta)^{1/2} d\vartheta
from 0 to \pi/3

Not sure how to integrate this.

You really made it hard on yourself.

r^{2} = 9 sin^2( \theta)
(dr/d\theta)^2 = (3 cos(\theta))^{2} = 9cos^2(\theta)
 

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